package com.health.mapp.utils;

/**
 * Tea算法 每次操作可以处理8个字节数�?KEY�?6字节,应为包含4个int型数的int[]，一个int�?个字�?
 * 加密解密轮数应为8的�?数，推荐加密轮数�?4�?
 * */
public class Tea {
	private final static int[] KEY = new int[] {// 加密解密�?��的KEY
	0x789f5645, 0xf68bd5a4, 0x81963ffa, 0x458fac58 };

	// 加密
	public byte[] encrypt(byte[] content, int offset, int[] key, int times) {// times为加密轮�?
		int[] tempInt = byteToInt(content, offset);
		int y = tempInt[0], z = tempInt[1], sum = 0, i;
		int delta = 0x9e3779b9; // 这是算法标准给的�?
		int a = key[0], b = key[1], c = key[2], d = key[3];

		for (i = 0; i < times; i++) {

			sum += delta;
			y += ((z << 4) + a) ^ (z + sum) ^ ((z >> 5) + b);
			z += ((y << 4) + c) ^ (y + sum) ^ ((y >> 5) + d);
		}
		tempInt[0] = y;
		tempInt[1] = z;
		return intToByte(tempInt, 0);
	}

	// 解密
	public byte[] decrypt(byte[] encryptContent, int offset, int[] key,
			int times) {
		int[] tempInt = byteToInt(encryptContent, offset);
		int y = tempInt[0], z = tempInt[1], sum = 0, i;
		int delta = 0x9e3779b9; // 这是算法标准给的�?
		int a = key[0], b = key[1], c = key[2], d = key[3];
		if (times == 32)
			sum = 0xC6EF3720; /* delta << 5 */
		else if (times == 16)
			sum = 0xE3779B90; /* delta << 4 */
		else
			sum = delta * times;

		for (i = 0; i < times; i++) {
			z -= ((y << 4) + c) ^ (y + sum) ^ ((y >> 5) + d);
			y -= ((z << 4) + a) ^ (z + sum) ^ ((z >> 5) + b);
			sum -= delta;
		}
		tempInt[0] = y;
		tempInt[1] = z;

		return intToByte(tempInt, 0);
	}

	// byte[]型数据转成int[]型数�?
	private int[] byteToInt(byte[] content, int offset) {

		int[] result = new int[content.length >> 2];// 除以2的n次方 == 右移n�?�?
													// content.length / 4 ==
													// content.length >> 2
		for (int i = 0, j = offset; j < content.length; i++, j += 4) {
			result[i] = transform(content[j + 3])
					| transform(content[j + 2]) << 8
					| transform(content[j + 1]) << 16 | (int) content[j] << 24;
		}
		return result;

	}

	// int[]型数据转成byte[]型数�?
	private byte[] intToByte(int[] content, int offset) {
		byte[] result = new byte[content.length << 2];// 乘以2的n次方 == 左移n�?�?
														// content.length * 4 ==
														// content.length << 2
		for (int i = 0, j = offset; j < result.length; i++, j += 4) {
			result[j + 3] = (byte) (content[i] & 0xff);
			result[j + 2] = (byte) ((content[i] >> 8) & 0xff);
			result[j + 1] = (byte) ((content[i] >> 16) & 0xff);
			result[j] = (byte) ((content[i] >> 24) & 0xff);
		}
		return result;
	}

	// 若某字节为负数则�?��其转成无符号正数
	private static int transform(byte temp) {
		int tempInt = (int) temp;
		if (tempInt < 0) {
			tempInt += 256;
		}
		return tempInt;
	}

}
